# Given $A$, find invertible $B$ such that $B^{-1}AB$ is positive

Given $A \in Mat(n,n,\mathbb R)$, is there always an invertible matrix B, such that $B^{-1}AB$ is positive, assuming all eigenvalues of A are positive and simple ?

If yes, is it possible to classify all invertible matrices B with this property ?

I tried hard to work with the perron-frobenius-theorem, but the problem is reversed. The perron-frobenius-theorem starts with a positive matrix and ensures at least one simple strictly positive eigenvalue. So, I cannot see how I can use it for my problem because I want to get a positive matrix starting with a non-positive one.

Additional question : Is there a special name or a notation for matrices with strict positive entries ?

• Consider $n=1$, $A =-1$. Commented Jul 12, 2014 at 22:31
• If $A$ has only negative eigenvalues, its trace and the trace of all similar matrices (that is of the form $B^{-1}AB$) will be negative. So negative entries are guaranteed. Commented Jul 12, 2014 at 22:36
• As for your other question, if you say "positive matrix", I think it is implied that you mean that all entries are positive. (Not to be confused with "positive definite" or a "positive determinant".) Furthermore "positive" implies "strict positive". Commented Jul 12, 2014 at 22:43
• With your new assumptions, the answer is yes, because of each eigenvalues is simple, then $A$ can be diagonalised, I.e. $BAB^{-1} = \rm{diag}(\lambda_1 , \dots, \lambda_n}$. Commented Jul 12, 2014 at 22:52
• But the diagonal matrix is still not positive. Commented Jul 12, 2014 at 22:56

The answer is positive. In dimension $n=1$ it's obvious, so from now $n\geq 2$. Let $\lambda_1<\lambda_2<\ldots<\lambda_n$ be the eigenvalues of $A$ , and let $X$ invertible such that $X^{-1}AX=diag(\lambda_1,\ldots,\lambda_n):=D$. Now consider the matrix $$U=\begin{vmatrix}1&-x&0&\ldots &0\\ 1&0&-x&\ldots &0\\ \vdots & 0 & 0 &\ldots &-x\\ 1&1&1&\ldots &1 \end{vmatrix}$$ where $x\geq 1$. $U$ is invertible, and apart for a positive factor we have $$U^{-1}=\begin{vmatrix} x&x&x&\ldots&x^2\\ -x-k&1&\ldots&1&x\\ 1&-x-k&\ldots&1&x\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 1&1&\ldots&-x-k&x \end{vmatrix}$$ where $k=n-2$. Let's show that, for a sufficient large $x$, $U^{-1}D\,U$ is positive (i will use $c$ to denote terms independent of x) $$U^{-1}DU\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}=\begin{pmatrix}\lambda_1x+\ldots+\lambda_{n-1}x+\lambda_nx^2\\x(\lambda_n-\lambda_1)+c\\x(\lambda_n-\lambda_2)+c\\\ldots\end{pmatrix}$$ $$U^{-1}DU\begin{pmatrix}0\\1\\\vdots\\0\\0\end{pmatrix}=\begin{pmatrix}x^2(\lambda_n-\lambda_1)\\x^2\lambda_1+x\lambda_n+c\\x(\lambda_n-\lambda_1)+c\\\vdots\\x(\lambda_n-\lambda_1)+c\end{pmatrix}$$ and so on. It's easy to see that, for sufficinet large $x$, all these vectors are positive. Now, set $B=XU$, then $B^{-1}AB$ is positive.